From: Mike Rosing <rosing@neurophys.wisc.edu>
Newsgroups: sci.physics.plasma
Subject: Re: Photon frequency loss on collision
Organization: Medical Electronics Lab.
References: <a1sloi$bvph$1@saturn.cs.uml.edu>


hanson wrote:
>
> Photon frequency loss on collision
>
> Does a photon loose energy when colliding with a H-atom, in a very dilute
> gas, like in intergalactic space?

It depends on the absorbtion cross section. If it's at some kind of
resonance,
then it could lose a lot of energy.

> How much, what fraction, of its incident frequency does a photon lose when
> encountering and colliding with an H atom?

It depends. What is the intial state of the H atom?

> It is said that there are all kinds of interactions possible from
> ionization to transnational, well, translational, and rotational states.
> But I don't want to have a rehash of Bohr, Heisenberg and Moessbauer
> theories.
>
> What I like to know is whether there is (perhaps) a common loss ratio of
> energy between incidence and emission of delta f / f, no matter what the
> initial hf of the photon is?

You mean a closed form simple formula? I'm sure it's possible to write
down something reasonable, but it won't account for resonances or the
higher energy starting points an H atom can be in.

> In case one insists hat hf-in = hf out, that NO frequency loss occurs
> between absorption and emission, then the question arises: What does the
> work, what makes this (unit=1) process happen?

That's a scattering interaction. The EM field of the photon interacts
with the E field of the electron and proton. If there is really no
change
in the photon, there is no interaction. You can transfer a very tiny
amount of energy between the incoming photon and the H atom, and to
good accuracy it's close to the same energy. But if you want *no*
change in energy of the photon, then can't be any interaction.

Patience, persistence, truth,
Dr. mike